what means starting voltage in solenoid valve?

Thread Starter

mohammad2050

Joined Nov 14, 2014
56
i want to drive a pneumatic ejector(solenoid valve), is specifications in following:

Starting Voltage : 100V
Maintaining voltage : 12V
Resistance Value :18Ω
Frequency(Hz):500-800
Maximum instantaneous power : 200W
Standby power : 3W​

what means starting voltage ? i know that i need to build a magnetic field across a gap. Then the gap closes so i need much less power (12 volt) to hold it in place. And 0V to let it turn off again but 100 volt is where? and how build circuit for drive?

do is below circuit appropriate ? if yes, 100 volt is where and how can obtain values?
 

dl324

Joined Mar 30, 2015
8,869
do is below circuit appropriate ?
The way I interpret the specs; no. The solenoid requires 100V (which seems high) to turn on and 12V to hold (it on). Lowering the voltage on the solenoid coil after it has engaged is to reduce power consumption.
 

MikeML

Joined Oct 2, 2009
5,444
Do you have a data sheet or link to one?

This solenoid is the type that needs to be pulsed (initially) with a high voltage, and then switched to a lower voltage to hold it in. If the high voltage is applied too long, the solenoid burns up due to excessive power dissipation...
 

dl324

Joined Mar 30, 2015
8,869
The obvious solution is to put a resistor paralleled with a capacitor in series with the solenoid coil. The RC time constant needs to be long enough for the solenoid to pull in.

The problem will be discharging the capacitor if the solenoid is switched quickly or you want it to be able to respond to momentary power glitches.
 

MikeML

Joined Oct 2, 2009
5,444
Two ways:

100Vdc power supply, NFET switch which is turned on for a few tens of msec.
12Vdc supply, NFET switch which is turned on after the one above turns off.


Other way: 100V supply and a switched constant-current driver. The voltage across the solenoid will drop to a steady-state value of I = E/R = 12/18 = 0.67A.
 

Thread Starter

mohammad2050

Joined Nov 14, 2014
56
Two ways:

100Vdc power supply, NFET switch which is turned on for a few tens of msec.
12Vdc supply, NFET switch which is turned on after the one above turns off.


Other way: 100V supply and a switched constant-current driver. The voltage across the solenoid will drop to a steady-state value of I = E/R = 12/18 = 0.67A.
ok , in your opinion do this circuit can doing 500-800 switchs in second?
 

MikeML

Joined Oct 2, 2009
5,444
What is the inductance of the coil? Is this some sort of proportional valve?

To actively drive it both open and closed at that rate will be very difficult. It will require split + and - 100V power supplies, and very powerful, very expensive, high voltage, high power opamp, like these: See PB63.
 

MaxHeadRoom

Joined Jul 18, 2013
19,006
If it is pneumatic it is very unusual to be a proportional type, these are usually hydraulic for accuracy.
Can you post a pic of the solenoid?
Max.
 

MikeML

Joined Oct 2, 2009
5,444
100V^2/18ohms = 555Watts
No! the inductance of the coil limits the power dissipation during the initial period when the 100V supply is first applied. You apply 100V, the current builds up. When the coil current reaches its steady-state limit (12V/18Ω = 0.67A), you switch from an applied 100V to 12V.
 

MikeML

Joined Oct 2, 2009
5,444
@mohammad2050 ,

Are you planning to use pwm? if so, what pulse frequency?

This will likely need a bipolar, constant-current driver with split +-100V supplies. In other words, you need to actively drive the coil with both current polarities.

You could use a single 100V supply, and an constant-current H-Bridge to reverse the current in the coil.
 

MaxHeadRoom

Joined Jul 18, 2013
19,006
If proportional, where is the feedback device, do all apertures activate at one time?
It appears to be just an on/off control?
It is still not clear on the mechanics of the device.
Max.
 
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